Question

If $\cot \theta -\sin \theta =\frac{1}{5}$ where $0<\theta <\frac{\pi }{2}$

Answer 1

$a\to q;b\to r;c\to s;d\to p;$

$\cos \theta -\sin \theta =\frac{1}{5}$,................(i)

where $0<\theta <\frac{\pi }{2}$

Squaring both sides of Eq. (i), we get

$1-\sin 2\theta =\frac{1}{25}$

$\Rightarrow \sin 2\theta =\frac{24}{25}\Rightarrow \cos 2\theta =\frac{7}{25}$

Also, ${(\cos \theta +\sin \theta )}^2={(\cos \theta -\sin \theta )}^2+4\cos \theta \sin \theta =\frac{1}{25}+2\sin 2\theta =\frac{1}{25}+\frac{48}{25}=\frac{49}{25}$

$\Rightarrow \cos \theta +\sin \theta =\frac{7}{5}$................(ii)

Also solving Eqs. (i) and (ii), we get $\cos \theta =4/5$